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Q.
Find the maximum/minimum value of the quadratic expression $3 x^2+10 x+8$.
Quadratic Equations
Solution:
Given: $x^2+6 x+12$
which is in the form of $a x^2+b x+c$.
where $a=1 >0$
$ \text { where } a=1 >0 $
$ \text { Minimum value }=\frac{4 a c-b^2}{4 a} $
$=\frac{4(3)(12)-36}{4(3)} $
$=\frac{48-36}{4}=3$